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3 years ago

Jill has two hamsters Each hamster drinks 3 ounces of water each day.

Mathematics

1 answer:

sveta [45]3 years ago

8 0

Answer:

4 oz

Step-by-step explanation:

2 hamsters 3 ounces each per day

6 ounces per day is drank

Monday 6 oz Tuesday 6 oz

16-12=4

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Mattias gets dressed in the dark one morning and chooses his clothes at random. He chooses a shirt (green, red, or yellow), a pa

Oliga [24]

Answer:

So, the number of ways in which the dress can be chosen

= 3 × 2 × 2

= 12.

And the combinations are

(Green shirt, black pant, checkered shoes),

(Green shirt, black pant, red shoes),

(Green shirt, blue pant, checkered shoes),

(Green shirt, blue pant, red shoes),

(Red shirt, black pant, checkered shoes),

(Red shirt, black pant, red shoes),

(Red shirt, blue pant, checkered shoes),

(Red shirt, blue pant, red shoes),

(Yellow shirt, black pant, checkered shoes),

(Yellow shirt, black pant, red shoes),

(Yellow shirt, blue pant, checkered shoes),

(Yellow shirt, blue pant, red shoes).

The tree diagram is below.

Step-by-step explanation:

7 0

3 years ago

Isaac is purchasing two pairs of shoes—one pair for $37.00 and the second pair for $42.00. The state sales tax applied to Isaac’

aleksandr82 [10.1K]

Answer:
7.11 and 9.48

Step-by-step explanation:

.12 is equal to 7.11
with out coupon

.12 is equal to 9.48
with the coupon

7 0

3 years ago

Relative extrema of f(x)=(x+3)/(x-2)

Salsk061 [2.6K]

Answer:

$\displaystyle f(x) = \frac{x + 3}{x - 2}$ has no relative extrema when the domain is $\mathbb{R} \backslash \lbrace 2 \rbrace$ (the set of all real numbers other than $2$ .)

Step-by-step explanation:

Assume that the domain of $\displaystyle f(x) = \frac{x + 3}{x - 2}$ is $\mathbb{R} \backslash \lbrace 2 \rbrace$ (the set of all real numbers other than $2$ .)

Let $f^{\prime}(x)$ and $f^{\prime\prime}(x)$ denote the first and second derivative of this function at $x$ .

Since this domain is an open interval, $x = a$ is a relative extremum of this function if and only if $f^{\prime}(a) = 0$ and $f^{\prime\prime}(a) \ne 0$ .

Hence, if it could be shown that $f^{\prime}(x) \ne 0$ for all $x \in \mathbb{R} \backslash \lbrace 2 \rbrace$ , one could conclude that it is impossible for $\displaystyle f(x) = \frac{x + 3}{x - 2}$ to have any relative extrema over this domain- regardless of the value of $f^{\prime\prime}(x)$ .

$\displaystyle f(x) = \frac{x + 3}{x - 2} = (x + 3) \, (x - 2)^{-1}$ .

Apply the product rule and the power rule to find $f^{\prime}(x)$ .

$\begin{aligned}f^{\prime}(x) &= \frac{d}{dx} \left[ (x + 3) \, (x - 2)^{-1}\right] \\ &= \left(\frac{d}{dx}\, [(x + 3)]\right)\, (x - 2)^{-1} \\ &\quad\quad (x + 3)\, \left(\frac{d}{dx}\, [(x - 2)^{-1}]\right) \\ &= (x - 2)^{-1} \\ &\quad\quad+ (x + 3) \, \left[(-1)\, (x - 2)^{-2}\, \left(\frac{d}{dx}\, [(x - 2)]\right) \right] \\ &= \frac{1}{x - 2} + \frac{-(x+ 3)}{(x - 2)^{2}} \\ &= \frac{(x - 2) - (x + 3)}{(x - 2)^{2}} = \frac{-5}{(x - 2)^{2}}\end{aligned}$ .

In other words, $\displaystyle f^{\prime}(x) = \frac{-5}{(x - 2)^{2}}$ for all $x \in \mathbb{R} \backslash \lbrace 2 \rbrace$ .

Since the numerator of this fraction is a non-zero constant, $f^{\prime}(x) \ne 0$ for all $x \in \mathbb{R} \backslash \lbrace 2 \rbrace$ . (To be precise, $f^{\prime}(x) < 0$ for all $x \in \mathbb{R} \backslash \lbrace 2 \rbrace\!$ .)

Hence, regardless of the value of $f^{\prime\prime}(x)$ , the function $f(x)$ would have no relative extrema over the domain $x \in \mathbb{R} \backslash \lbrace 2 \rbrace$ .